Algebra - Circles

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Section 3.3 : Circles

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6. Determine the radius and center of the following circle. If the equation is not the equation of a circle clearly explain why not.

${x^2} + {y^2} + 14x - 8y + 56 = 0$

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To do this problem we need to complete the square on the $x$ and $y$ terms. To help with this we’ll first get the number on the right side and group the $x$ and $y$ terms as follows.

${x^2} + 14x + {y^2} - 8y = - 56$ Show Step 2

Here are the numbers we need to complete the square for both $x$ and $y$ .

$x:\,{\left( {\frac{{14}}{2}} \right)^2} = {\left( 7 \right)^2} = 49\hspace{0.5in}y:\, {\left( {\frac{{ - 8}}{2}} \right)^2} = {\left( { - 4} \right)^2} = 16$ Show Step 3

Now, complete the square.

$\begin{align*}{x^2} + 14x + 49 + {y^2} - 8y + 16 & = - 56 + 49 + 16\\ {\left( {x + 7} \right)^2} + {\left( {y - 4} \right)^2} & = 9\end{align*}$

Don’t forget to add the numbers from Step 2 to both sides of the equation!

Show Step 4

So, we have the equation in standard form and so we can quickly identify the radius and center of the circle.

${\mbox{Radius : }}r = 3\hspace{0.25in}\hspace{0.25in}{\mbox{Center : }}\left( { - 7,4} \right)$

If you don’t recall how to get the radius and center from the standard form of a circle check out Problems 3 – 5 in this section for some practice.